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Math Help - help with algebra in calc question: infinite limit

  1. #1
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    help with algebra in calc question: infinite limit

    i was trying to find the limits of a function when i came onto a problem

    square root of ( (x^2) + 9x + 3)
    multiply by (1/x)

    this should simplify to:

    square root of ((x) + (9/x) + (3 / (x^2) )

    how does this work?

    i was trying to slove this problem

    find the limit of

    ((square root of ((x^2) + 9x + 3)) minus x

    lim when x approaches infinity


    just convert the 1/x into square root of (1/ (x^2)
    this makes it possible to simplify the stuff under the main square root.
    this is why x^2 simplifies to 1, 9x simplifies to 9/x, etc.

    thanks to me?
    Last edited by Jhevon; March 11th 2009 at 06:19 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jedward View Post
    just convert the 1/x into square root of (1/ (x^2)
    this makes it possible to simplify the stuff under the main square root.
    this is why x^2 simplifies to 1, 9x simplifies to 9/x, etc.

    thanks to me?
    as it stands, you cannot multiply by 1/x, unless there is some part of the question you are not writing, in which case you can multiply by something like \frac {\frac 1x}{\frac 1x}, if it were a rational function, but it depends.

    \lim_{x \to \infty} \sqrt{x^2 + 9x + 3} = \infty that was your first question


    the second is different. you would multiply by the conjugate over itself to find the limit

    \lim_{x \to \infty} \sqrt{x^2 + 9x + 3} - x = \lim_{x \to \infty} ( \sqrt{x^2 + 9x + 3} - x) \cdot \frac {\sqrt{x^2 + 9x + 3} + x}{\sqrt{x^2 + 9x + 3} + x}



    now simplify and finish up. you will need to note the following: \sqrt{x^2 + 9x + 3} = \sqrt{x^2 \left(1 + \frac 9x + \frac 3{x^2} \right)} = \sqrt{x^2} \sqrt{1 + \frac 9x + \frac 3{x^2}} = |x| \sqrt{1 + \frac 9x + \frac 3{x^2}}
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  3. #3
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    Hello, jedward!

    \lim_{x\to\infty}\bigg[\sqrt{x^2+9x+3} - x\bigg]
    Multiply top and bottom by the conjugate:

    . . \frac{\sqrt{x^2+9x+3} - x}{1}\;\cdot\;\frac{\sqrt{x^2+9x+3} + x}{\sqrt{x^2+9x+3} + x} \;=\;\frac{(x^2+9x+3) - x^2}{\sqrt{x^2+9x+3} + x}  \;=\;\frac{9x+3}{\sqrt{x^2+9x+3} + x}


    Divide top and bottom by x:

    . . \frac{\dfrac{9x}{x} + \dfrac{3}{x}} { \dfrac{\sqrt{x^2+9x+3}}{x} + \dfrac{x}{x}} \;=\;\frac{9+\dfrac{3}{x}} {\dfrac{\sqrt{x^2+9x+3}}{\sqrt{x^2}} + 1} = \;\frac{9+\dfrac{3}{x}}{\sqrt{\dfrac{x^2+9x+3}{x^2  }} + 1}

    . . . . =\;\frac{9+\dfrac{3}{x}}{\sqrt{\dfrac{x^2}{x^2} + \dfrac{9x}{x^2} + \dfrac{3}{x^2}} + 1} \;=\;\frac{9+\dfrac{3}{x}}{\sqrt{1 + \dfrac{9}{x} + \dfrac{3}{x^2}} + 1}


    Therefore: . \lim_{x\to\infty}\left[\frac{9+\dfrac{3}{x}}{\sqrt{1 + \dfrac{9}{x} + \dfrac{3}{x^2}} + 1}\right] \;=\; \frac{9 + 0}{\sqrt{1+0+0} + 1} \;=\;\frac{9}{2}


    Edit: Jhevon beat me to it . . .
    .
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